haskell - 在 monad 转换器中，为什么已知的 monad 是内部的？

Question

例如，MaybeT定义为:

newtype MaybeT m a =
  MaybeT { runMaybeT :: m (Maybe a)}

但不是:

newtype MaybeT m a =
  MaybeT { runMaybeT :: Maybe (m a) }

为什么是这样？

Answer 1

扩展新类型后，我们有 join :: Monad m => m (Maybe (m (Maybe a))) -> m (Maybe a)在第一种情况下，join :: Monad m => Maybe (m (Maybe (m a))) -> Maybe (m a)在第二。

实现第一个join您需要一种分发方式Maybe超过 m : dist1 :: Monad m => Maybe (m a) -> m (Maybe a) :

join1 :: m (Maybe (m (Maybe a))) -> m (Maybe a)
join1 = fmap join . join . fmap dist1

实现第二个 join你需要相反的分配律 dist2 :: Monad m => m (Maybe a) -> Maybe (m a)

join2 :: Maybe (m (Maybe (m a))) -> Maybe (m a)
join2 = fmap join . join . fmap dist2

dist1很容易实现(我将证明 monad 更改器(mutator)定律留给你):

dist1 :: Monad m => Maybe (m a) -> m (Maybe a) 
dist1 = sequenceA

dist2不是那么容易。它不能用于任意 Monad .作为反例，我们选择 m成为“读者”monad (->) r :

dist2 :: (r -> Maybe a) -> Maybe (r -> a)

由于您无权访问 r ， dist2 的唯一实现那将是 const Nothing ，这显然不满足单子(monad)定律。